Why is it clear that $A\times(B\times C) = (B\wedge C)(L(A),\cdot)?$

Question

As motivation, this exercise claims to be a way to gain intuition for the $BAC-CAB$ identity, but I'm gaining a sum total of zero intuition by doing it.

First and foremost, I have to lay out notation. In the title of this post, $A,B,C\in\mathbb{R}^3$ and $L:\mathbb{R}^3\rightarrow\mathbb{R}^{3*}$. Furthermore, I'm using the dual metric for all of this: $L(A)(\cdot) = (A|\cdot)$. The empty slot in $(L(A),\cdot)$ is to emphasize that this is partial application of the $(0,2)$ tensor $B\wedge C$, defined by $(B\otimes C)(L(A),\cdot) = B(C(L(A))) = B(L(A)(C)) = B(A|C)$.

Now, having laid all of this out, and using the Euclidean metric $(A|B) = A\cdot B$, how am I supposed to see why $A\times(B\times C) = (B\wedge C)(L(A),\cdot)$? Carrying out the RHS, it's direct to show that

$$(B\wedge C)(L(A),\cdot) = (B\otimes C - C\otimes B)(L(A),\cdot) = B(A\cdot C) - C(A\cdot B)$$

thus yielding the familiar $BAC-CAB$ identity, but I fail to see why I would be able to make this association with that triple-cross in the first place. The route that would be clear to me would be the following. By considering an $\alpha\in\Lambda^2\mathbb{R}^3$ and defining $J:\Lambda^2\mathbb{R}^3\rightarrow\mathbb{R}^3$ to have it's i'th component as $(J(\alpha))^i = \frac{1}{2}\epsilon^i_{jk}\alpha^{jk}$, it follows that $A\times B = J(A\wedge B)$, so for this problem I would be more inclined to do

$$A\times(B\times C) = J(A\wedge J(B\wedge C))$$

and if, from there, I could somehow see that this is equivalent to $(B\wedge C)(L(A),\cdot)$, then I would gain the "intuition" that this exercise claims to be the motivation. Just showing both sides to be $BAC-CAB$ leaves me empty.

Answer 1

I don't know exactly what you're looking for, but here are my thoughts anyway. I'll use multiplication (as defined here) instead of function notation which I find confusing and unnecessary.

With the unit trivector $I=e_1e_2e_3$, the cross product can be defined as

$$A\times B=-(A\wedge B)I=-(A\wedge B)\,\lrcorner\,I=-A\,\lrcorner\,(B\,\lrcorner\,I)=-A\,\lrcorner\,(BI).$$

(I have here a proof of the middle equation.) Thus the triple product is

$$A\times(B\times C)=-A\,\lrcorner\,\Big(\big(-(B\wedge C)I\big)I\Big)=A\,\lrcorner\,\big((B\wedge C)I^2\big)=-A\,\lrcorner\,(B\wedge C).$$

Contraction of a vector with a bivector is anticommutative, so this is $(B\wedge C)\,\llcorner\, A$.

To see that this is $(A\cdot C)B-(A\cdot B)C$, I would just verify that it's true for the basis vectors $e_i$ (which is easier because they're orthogonal, and the products reduce to the geometric product or $0$), and then it's true for all vectors because the expressions are multilinear. Indeed, I used this method for most identities when I was first learning geometric algebra. But it may be unsatisfying.

Alternatively, we can break $A$ into parts parallel and perpendicular to the $B\wedge C$ plane, as in @calcstudent's answer; or write the products as symmetric and antisymmetric parts of the geometric product:

$$B\wedge C=\frac12(BC-CB)$$

$$(B\wedge C)\,\llcorner\,A=\frac12\big((B\wedge C)A-A(B\wedge C)\big)$$

$$=\frac14(BCA-CBA-ABC+ACB)$$

$$=\frac14\big(B(AC+CA)+(AC+CA)B-C(AB+BA)-(AB+BA)C\big)$$

$$=\frac12\big(B(A\cdot C)+(A\cdot C)B-C(A\cdot B)-(A\cdot B)C\big)$$

$$=(A\cdot C)B-(A\cdot B)C.$$

Answer 2

Write $B\times C$ as $D$, then $D$ is a blade. Let $I$ be the unit pseudoscalar for right-hand rule. Then $J(B\times C)=-ID$. The equation is now written as $A\times(-ID)=D(L(A),\cdot)$.

I think the interpretation here is that both side compute the same quantity. This quantity is a form of dot product between vector $A$ and blade $D$, which is the result of canceling out the component of $D$ the direction along $A$ leaving only the scalar, while retaining the direction orthogonal to $A$. But they do it in different manner: the left hand side do it by representing $D$ with a vector using the right hand rule representation, the right hand side do it by projecting $A$ down onto $D$ and canceling.

So now to prove it. The claim is clearly true if $D=0$ so assume not, so $D$ represent an unique plane. By linearity, it's sufficient to check the claim for: $A$ orthogonal to $D$ plane, and $A$ parallel to $D$ plane. If $A$ is orthogonal to $D$ then both side are trivially $0$. So we assume $A$ is parallel to $D$ and not orthogonal, or in other word $A$ lie on $D$ plane and is nonzero.

Let $u=\frac{A}{|A|}$ then we can write $D=uv=u\wedge v$ for some $v$ orthogonal to $u$ and is on the plane represented by $D$; specifically $v=uD$.

Then $-ID=u\times v$. So $A\times(-ID)=A\times(u\times v)=|A|u\times(u\times v)=-|A|v$ because left multiplication by unit vector $u$ induce a right angle rotation on the plane orthogonal to it according to right hand rule. And $(u\wedge v)(L(A),\cdot)=|A|(u\otimes v-v\otimes u)(u,\cdot)=|A|u(u|v)-|A|v(u|u)=-|A|v$

One key property we are using here is how flexible it is to represent a blade using vectors on it, which is why since the beginning we are treating this as an equation between a vector $A$ and a blade $D$, ignoring $B$ and $C$ completely. Once you take wedge product, the only thing that matter is what blade does $B$ and $C$ form and not they are in particular (and same for cross product which is the result of wedge product times unit pseudoscalar).